YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , plus(N, s(M)) -> U51(isNat(M), M, N) , plus(N, 0()) -> U41(isNat(N), N) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) , x(X1, X2) -> n__x(X1, X2) , x(N, s(M)) -> U71(isNat(M), M, N) , x(N, 0()) -> U61(isNat(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { plus(N, s(M)) -> U51(isNat(M), M, N) , plus(N, 0()) -> U41(isNat(N), N) , x(N, s(M)) -> U71(isNat(M), M, N) , x(N, 0()) -> U61(isNat(N)) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) , x(X1, X2) -> n__x(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { isNat(n__0()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [2] x2 + [0] [tt] = [0] [U12](x1) = [1] x1 + [0] [isNat](x1) = [2] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [2] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [7] [U51](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [4] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [7] x1 + [7] [0] = [4] [U71](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [4] [x](x1, x2) = [1] x1 + [1] x2 + [0] [n__0] = [4] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [0] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [2] V2 + [0] >= [2] V2 + [0] = [U12(isNat(activate(V2)))] [U12(tt())] = [0] >= [0] = [tt()] [isNat(n__0())] = [8] > [0] = [tt()] [isNat(n__plus(V1, V2))] = [2] V2 + [2] V1 + [0] >= [2] V2 + [2] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [2] V1 + [0] >= [2] V1 + [0] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [2] V2 + [2] V1 + [0] >= [2] V2 + [2] V1 + [0] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [4] >= [4] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [x(X1, X2)] [U21(tt())] = [0] >= [0] = [tt()] [U31(tt(), V2)] = [2] V2 + [0] >= [2] V2 + [0] = [U32(isNat(activate(V2)))] [U32(tt())] = [0] >= [0] = [tt()] [U41(tt(), N)] = [7] N + [7] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [4] M + [0] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [4] N + [4] M + [0] >= [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [7] > [4] = [0()] [0()] = [4] >= [4] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [4] N + [2] M + [4] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [4] > [2] N + [1] M + [0] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { isNat(n__0()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { U31(tt(), V2) -> U32(isNat(activate(V2))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [0] [U12](x1) = [1] x1 + [0] [isNat](x1) = [1] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [1] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [7] [U51](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [2] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [7] x1 + [7] [0] = [0] [U71](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [2] x2 + [2] x3 + [4] [x](x1, x2) = [1] x1 + [1] x2 + [1] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [1] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [0] >= [1] V2 + [0] = [U12(isNat(activate(V2)))] [U12(tt())] = [0] >= [0] = [tt()] [isNat(n__0())] = [0] >= [0] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [0] >= [1] V2 + [1] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [0] >= [1] V1 + [0] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [1] >= [1] V2 + [1] V1 + [1] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [x(X1, X2)] [U21(tt())] = [0] >= [0] = [tt()] [U31(tt(), V2)] = [1] V2 + [1] > [1] V2 + [0] = [U32(isNat(activate(V2)))] [U32(tt())] = [0] >= [0] = [tt()] [U41(tt(), N)] = [7] N + [7] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [7] > [3] N + [4] M + [0] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [2] N + [4] M + [0] >= [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [7] > [0] = [0()] [0()] = [0] >= [0] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [4] N + [2] M + [4] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [4] > [2] N + [1] M + [1] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U32(tt()) -> tt() , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { isNat(n__0()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { U52(tt(), M, N) -> s(plus(activate(N), activate(M))) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [1] [U12](x1) = [1] x1 + [0] [isNat](x1) = [1] x1 + [1] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [7] [U51](x1, x2, x3) = [1] x1 + [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [1] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [7] x1 + [7] [0] = [0] [U71](x1, x2, x3) = [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [4] x1 + [1] x2 + [2] x3 + [0] [x](x1, x2) = [1] x1 + [1] x2 + [0] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [0] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [1] >= [1] V2 + [1] = [U12(isNat(activate(V2)))] [U12(tt())] = [1] >= [1] = [tt()] [isNat(n__0())] = [1] >= [1] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [1] >= [1] V2 + [1] V1 + [1] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [1] >= [1] V1 + [1] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [1] >= [1] V2 + [1] V1 + [1] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [x(X1, X2)] [U21(tt())] = [1] >= [1] = [tt()] [U31(tt(), V2)] = [1] V2 + [1] >= [1] V2 + [1] = [U32(isNat(activate(V2)))] [U32(tt())] = [1] >= [1] = [tt()] [U41(tt(), N)] = [7] N + [14] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [8] > [2] N + [4] M + [1] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [4] M + [1] > [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [14] > [0] = [0()] [0()] = [0] >= [0] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [1] M + [4] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [4] > [2] N + [1] M + [0] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U32(tt()) -> tt() , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { isNat(n__0()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [4] x2 + [0] [tt] = [0] [U12](x1) = [1] x1 + [0] [isNat](x1) = [4] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [4] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [7] [U51](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [3] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [1] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [7] x1 + [7] [0] = [0] [U71](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [1] x1 + [1] x2 + [2] x3 + [4] [x](x1, x2) = [1] x1 + [1] x2 + [1] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [1] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [4] V2 + [0] >= [4] V2 + [0] = [U12(isNat(activate(V2)))] [U12(tt())] = [0] >= [0] = [tt()] [isNat(n__0())] = [0] >= [0] = [tt()] [isNat(n__plus(V1, V2))] = [4] V2 + [4] V1 + [0] >= [4] V2 + [4] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [4] V1 + [0] >= [4] V1 + [0] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [4] V2 + [4] V1 + [4] > [4] V2 + [4] V1 + [0] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [x(X1, X2)] [U21(tt())] = [0] >= [0] = [tt()] [U31(tt(), V2)] = [4] V2 + [0] >= [4] V2 + [0] = [U32(isNat(activate(V2)))] [U32(tt())] = [0] >= [0] = [tt()] [U41(tt(), N)] = [7] N + [7] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [3] > [5] N + [4] M + [0] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [4] M + [0] >= [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [7] > [0] = [0()] [0()] = [0] >= [0] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [1] M + [4] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [4] > [2] N + [1] M + [1] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U32(tt()) -> tt() , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { isNat(n__0()) -> tt() , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , U31(tt(), V2) -> U32(isNat(activate(V2))) , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [2] x2 + [0] [tt] = [4] [U12](x1) = [1] x1 + [1] [isNat](x1) = [2] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [2] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [3] x1 + [7] x2 + [3] [U51](x1, x2, x3) = [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [4] x3 + [0] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [3] x1 + [3] [0] = [4] [U71](x1, x2, x3) = [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [2] x2 + [2] x3 + [0] [x](x1, x2) = [1] x1 + [1] x2 + [0] [n__0] = [4] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [0] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [2] V2 + [4] > [2] V2 + [1] = [U12(isNat(activate(V2)))] [U12(tt())] = [5] > [4] = [tt()] [isNat(n__0())] = [8] > [4] = [tt()] [isNat(n__plus(V1, V2))] = [2] V2 + [2] V1 + [0] >= [2] V2 + [2] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [2] V1 + [0] >= [2] V1 + [0] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [2] V2 + [2] V1 + [0] >= [2] V2 + [2] V1 + [0] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [4] >= [4] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [x(X1, X2)] [U21(tt())] = [4] >= [4] = [tt()] [U31(tt(), V2)] = [2] V2 + [4] > [2] V2 + [0] = [U32(isNat(activate(V2)))] [U32(tt())] = [4] >= [4] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [4] M + [0] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [4] N + [4] M + [4] > [1] N + [1] M + [0] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [15] > [4] = [0()] [0()] = [4] >= [4] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [2] M + [0] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [8] > [2] N + [1] M + [0] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U32(tt()) -> tt() , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , U31(tt(), V2) -> U32(isNat(activate(V2))) , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [4] x2 + [0] [tt] = [0] [U12](x1) = [1] x1 + [0] [isNat](x1) = [4] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [4] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [7] [U51](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [3] [U52](x1, x2, x3) = [1] x1 + [1] x2 + [2] x3 + [1] [s](x1) = [1] x1 + [0] [plus](x1, x2) = [1] x1 + [1] x2 + [1] [U61](x1) = [7] x1 + [7] [0] = [0] [U71](x1, x2, x3) = [7] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [1] x1 + [2] x2 + [2] x3 + [4] [x](x1, x2) = [1] x1 + [1] x2 + [0] [n__0] = [0] [n__plus](x1, x2) = [1] x1 + [1] x2 + [1] [n__s](x1) = [1] x1 + [0] [n__x](x1, x2) = [1] x1 + [1] x2 + [0] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [4] V2 + [0] >= [4] V2 + [0] = [U12(isNat(activate(V2)))] [U12(tt())] = [0] >= [0] = [tt()] [isNat(n__0())] = [0] >= [0] = [tt()] [isNat(n__plus(V1, V2))] = [4] V2 + [4] V1 + [4] > [4] V2 + [4] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [4] V1 + [0] >= [4] V1 + [0] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [4] V2 + [4] V1 + [0] >= [4] V2 + [4] V1 + [0] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [0] >= [0] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [x(X1, X2)] [U21(tt())] = [0] >= [0] = [tt()] [U31(tt(), V2)] = [4] V2 + [0] >= [4] V2 + [0] = [U32(isNat(activate(V2)))] [U32(tt())] = [0] >= [0] = [tt()] [U41(tt(), N)] = [7] N + [7] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [3] > [6] N + [1] M + [1] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [2] N + [1] M + [1] >= [1] N + [1] M + [1] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [1] >= [1] X1 + [1] X2 + [1] = [n__plus(X1, X2)] [U61(tt())] = [7] > [0] = [0()] [0()] = [0] >= [0] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [2] M + [4] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [4] > [2] N + [1] M + [1] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { isNat(n__s(V1)) -> U21(isNat(activate(V1))) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U32(tt()) -> tt() , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , U31(tt(), V2) -> U32(isNat(activate(V2))) , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { isNat(n__s(V1)) -> U21(isNat(activate(V1))) , U21(tt()) -> tt() , U32(tt()) -> tt() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [2] x2 + [0] [tt] = [5] [U12](x1) = [1] x1 + [0] [isNat](x1) = [2] x1 + [0] [activate](x1) = [1] x1 + [0] [U21](x1) = [1] x1 + [3] [U31](x1, x2) = [1] x1 + [2] x2 + [4] [U32](x1) = [1] x1 + [4] [U41](x1, x2) = [3] x1 + [7] x2 + [0] [U51](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [1] [U52](x1, x2, x3) = [1] x1 + [2] x2 + [1] x3 + [0] [s](x1) = [1] x1 + [2] [plus](x1, x2) = [1] x1 + [1] x2 + [0] [U61](x1) = [3] x1 + [0] [0] = [4] [U71](x1, x2, x3) = [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [2] x2 + [2] x3 + [2] [x](x1, x2) = [1] x1 + [1] x2 + [4] [n__0] = [4] [n__plus](x1, x2) = [1] x1 + [1] x2 + [0] [n__s](x1) = [1] x1 + [2] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [2] V2 + [5] > [2] V2 + [0] = [U12(isNat(activate(V2)))] [U12(tt())] = [5] >= [5] = [tt()] [isNat(n__0())] = [8] > [5] = [tt()] [isNat(n__plus(V1, V2))] = [2] V2 + [2] V1 + [0] >= [2] V2 + [2] V1 + [0] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [2] V1 + [4] > [2] V1 + [3] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [2] V2 + [2] V1 + [8] > [2] V2 + [2] V1 + [4] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__0())] = [4] >= [4] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [2] >= [1] X + [2] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [x(X1, X2)] [U21(tt())] = [8] > [5] = [tt()] [U31(tt(), V2)] = [2] V2 + [9] > [2] V2 + [4] = [U32(isNat(activate(V2)))] [U32(tt())] = [9] > [5] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [0] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [11] > [3] N + [2] M + [0] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [2] M + [5] > [1] N + [1] M + [2] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [2] >= [1] X + [2] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__plus(X1, X2)] [U61(tt())] = [15] > [4] = [0()] [0()] = [4] >= [4] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [7] > [6] N + [2] M + [2] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [12] > [2] N + [1] M + [4] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , plus(X1, X2) -> n__plus(X1, X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [2] [U12](x1) = [1] x1 + [0] [isNat](x1) = [1] x1 + [0] [activate](x1) = [1] x1 + [1] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [7] x1 + [7] x2 + [1] [U51](x1, x2, x3) = [4] x1 + [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [4] x1 + [4] x2 + [3] x3 + [2] [s](x1) = [1] x1 + [4] [plus](x1, x2) = [1] x1 + [1] x2 + [4] [U61](x1) = [7] x1 + [1] [0] = [4] [U71](x1, x2, x3) = [3] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [1] x2 + [2] x3 + [7] [x](x1, x2) = [1] x1 + [1] x2 + [4] [n__0] = [4] [n__plus](x1, x2) = [1] x1 + [1] x2 + [3] [n__s](x1) = [1] x1 + [4] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [2] > [1] V2 + [1] = [U12(isNat(activate(V2)))] [U12(tt())] = [2] >= [2] = [tt()] [isNat(n__0())] = [4] > [2] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [3] > [1] V2 + [1] V1 + [2] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [4] > [1] V1 + [1] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [4] > [1] V2 + [1] V1 + [2] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [1] > [1] X + [0] = [X] [activate(n__0())] = [5] > [4] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [5] > [1] X + [4] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [5] > [1] X1 + [1] X2 + [4] = [x(X1, X2)] [U21(tt())] = [2] >= [2] = [tt()] [U31(tt(), V2)] = [1] V2 + [2] > [1] V2 + [1] = [U32(isNat(activate(V2)))] [U32(tt())] = [2] >= [2] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [1] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [15] > [7] N + [4] M + [13] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [3] N + [4] M + [10] >= [1] N + [1] M + [10] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [4] >= [1] X + [4] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [3] = [n__plus(X1, X2)] [U61(tt())] = [15] > [4] = [0()] [0()] = [4] >= [4] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [13] > [4] N + [1] M + [12] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [11] >= [2] N + [1] M + [11] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { activate(n__plus(X1, X2)) -> plus(X1, X2) , s(X) -> n__s(X) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { activate(n__plus(X1, X2)) -> plus(X1, X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [7] [U12](x1) = [1] x1 + [1] [isNat](x1) = [1] x1 + [0] [activate](x1) = [1] x1 + [2] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [0] [U32](x1) = [1] x1 + [1] [U41](x1, x2) = [2] x1 + [7] x2 + [1] [U51](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [1] [U52](x1, x2, x3) = [1] x1 + [4] x2 + [1] x3 + [3] [s](x1) = [1] x1 + [2] [plus](x1, x2) = [1] x1 + [1] x2 + [4] [U61](x1) = [2] x1 + [1] [0] = [7] [U71](x1, x2, x3) = [1] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [2] x2 + [2] x3 + [0] [x](x1, x2) = [1] x1 + [1] x2 + [4] [n__0] = [7] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__s](x1) = [1] x1 + [2] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [7] > [1] V2 + [3] = [U12(isNat(activate(V2)))] [U12(tt())] = [8] > [7] = [tt()] [isNat(n__0())] = [7] >= [7] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [2] >= [1] V1 + [2] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [2] > [1] X + [0] = [X] [activate(n__0())] = [9] > [7] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [4] > [1] X + [2] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [x(X1, X2)] [U21(tt())] = [7] >= [7] = [tt()] [U31(tt(), V2)] = [1] V2 + [7] > [1] V2 + [3] = [U32(isNat(activate(V2)))] [U32(tt())] = [8] > [7] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [2] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [15] >= [2] N + [4] M + [15] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [4] M + [10] >= [1] N + [1] M + [10] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [2] >= [1] X + [2] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [U61(tt())] = [15] > [7] = [0()] [0()] = [7] >= [7] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [14] > [4] N + [2] M + [12] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [2] M + [14] >= [2] N + [1] M + [14] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { s(X) -> n__s(X) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { s(X) -> n__s(X) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [5] [U12](x1) = [1] x1 + [0] [isNat](x1) = [1] x1 + [0] [activate](x1) = [1] x1 + [2] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [3] x1 + [7] x2 + [0] [U51](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [5] [U52](x1, x2, x3) = [2] x1 + [1] x2 + [1] x3 + [4] [s](x1) = [1] x1 + [3] [plus](x1, x2) = [1] x1 + [1] x2 + [4] [U61](x1) = [3] x1 + [0] [0] = [6] [U71](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [5] [U72](x1, x2, x3) = [2] x1 + [1] x2 + [2] x3 + [4] [x](x1, x2) = [1] x1 + [1] x2 + [4] [n__0] = [6] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__s](x1) = [1] x1 + [2] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [5] > [1] V2 + [2] = [U12(isNat(activate(V2)))] [U12(tt())] = [5] >= [5] = [tt()] [isNat(n__0())] = [6] > [5] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [2] >= [1] V1 + [2] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [2] > [1] X + [0] = [X] [activate(n__0())] = [8] > [6] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [4] > [1] X + [3] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [x(X1, X2)] [U21(tt())] = [5] >= [5] = [tt()] [U31(tt(), V2)] = [1] V2 + [5] > [1] V2 + [2] = [U32(isNat(activate(V2)))] [U32(tt())] = [5] >= [5] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [2] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [15] > [3] N + [1] M + [12] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [1] M + [14] > [1] N + [1] M + [11] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [3] > [1] X + [2] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [U61(tt())] = [15] > [6] = [0()] [0()] = [6] >= [6] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [15] > [4] N + [1] M + [14] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [14] >= [2] N + [1] M + [14] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { 0() -> n__0() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [1] x2 + [0] [tt] = [6] [U12](x1) = [1] x1 + [0] [isNat](x1) = [1] x1 + [0] [activate](x1) = [1] x1 + [2] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [1] x2 + [0] [U32](x1) = [1] x1 + [0] [U41](x1, x2) = [2] x1 + [7] x2 + [3] [U51](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [3] [U52](x1, x2, x3) = [2] x1 + [4] x2 + [1] x3 + [0] [s](x1) = [1] x1 + [2] [plus](x1, x2) = [1] x1 + [1] x2 + [4] [U61](x1) = [2] x1 + [3] [0] = [7] [U71](x1, x2, x3) = [1] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [1] x2 + [2] x3 + [2] [x](x1, x2) = [1] x1 + [1] x2 + [4] [n__0] = [6] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__s](x1) = [1] x1 + [2] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [1] V2 + [6] > [1] V2 + [2] = [U12(isNat(activate(V2)))] [U12(tt())] = [6] >= [6] = [tt()] [isNat(n__0())] = [6] >= [6] = [tt()] [isNat(n__plus(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [1] V1 + [2] >= [1] V1 + [2] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [1] V2 + [1] V1 + [4] >= [1] V2 + [1] V1 + [4] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [2] > [1] X + [0] = [X] [activate(n__0())] = [8] > [7] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [4] > [1] X + [2] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [6] > [1] X1 + [1] X2 + [4] = [x(X1, X2)] [U21(tt())] = [6] >= [6] = [tt()] [U31(tt(), V2)] = [1] V2 + [6] > [1] V2 + [2] = [U32(isNat(activate(V2)))] [U32(tt())] = [6] >= [6] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [2] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [15] > [3] N + [4] M + [14] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [4] M + [12] > [1] N + [1] M + [10] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [2] >= [1] X + [2] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [U61(tt())] = [15] > [7] = [0()] [0()] = [7] > [6] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [13] > [4] N + [1] M + [12] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [14] >= [2] N + [1] M + [14] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { x(X1, X2) -> n__x(X1, X2) } Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { x(X1, X2) -> n__x(X1, X2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(U11) = {1, 2}, Uargs(U12) = {1}, Uargs(isNat) = {1}, Uargs(U21) = {1}, Uargs(U31) = {1, 2}, Uargs(U32) = {1}, Uargs(U52) = {1, 2, 3}, Uargs(s) = {1}, Uargs(plus) = {1, 2}, Uargs(U72) = {1, 2, 3}, Uargs(x) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [U11](x1, x2) = [1] x1 + [2] x2 + [0] [tt] = [4] [U12](x1) = [1] x1 + [2] [isNat](x1) = [2] x1 + [0] [activate](x1) = [1] x1 + [1] [U21](x1) = [1] x1 + [0] [U31](x1, x2) = [1] x1 + [2] x2 + [4] [U32](x1) = [1] x1 + [1] [U41](x1, x2) = [3] x1 + [7] x2 + [3] [U51](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [7] [U52](x1, x2, x3) = [2] x1 + [2] x2 + [1] x3 + [4] [s](x1) = [1] x1 + [4] [plus](x1, x2) = [1] x1 + [1] x2 + [4] [U61](x1) = [3] x1 + [3] [0] = [5] [U71](x1, x2, x3) = [2] x1 + [7] x2 + [7] x3 + [7] [U72](x1, x2, x3) = [2] x1 + [1] x2 + [2] x3 + [5] [x](x1, x2) = [1] x1 + [1] x2 + [5] [n__0] = [4] [n__plus](x1, x2) = [1] x1 + [1] x2 + [4] [n__s](x1) = [1] x1 + [4] [n__x](x1, x2) = [1] x1 + [1] x2 + [4] The following symbols are considered usable {U11, U12, isNat, activate, U21, U31, U32, U41, U51, U52, s, plus, U61, 0, U71, U72, x} The order satisfies the following ordering constraints: [U11(tt(), V2)] = [2] V2 + [4] >= [2] V2 + [4] = [U12(isNat(activate(V2)))] [U12(tt())] = [6] > [4] = [tt()] [isNat(n__0())] = [8] > [4] = [tt()] [isNat(n__plus(V1, V2))] = [2] V2 + [2] V1 + [8] > [2] V2 + [2] V1 + [4] = [U11(isNat(activate(V1)), activate(V2))] [isNat(n__s(V1))] = [2] V1 + [8] > [2] V1 + [2] = [U21(isNat(activate(V1)))] [isNat(n__x(V1, V2))] = [2] V2 + [2] V1 + [8] >= [2] V2 + [2] V1 + [8] = [U31(isNat(activate(V1)), activate(V2))] [activate(X)] = [1] X + [1] > [1] X + [0] = [X] [activate(n__0())] = [5] >= [5] = [0()] [activate(n__plus(X1, X2))] = [1] X1 + [1] X2 + [5] > [1] X1 + [1] X2 + [4] = [plus(X1, X2)] [activate(n__s(X))] = [1] X + [5] > [1] X + [4] = [s(X)] [activate(n__x(X1, X2))] = [1] X1 + [1] X2 + [5] >= [1] X1 + [1] X2 + [5] = [x(X1, X2)] [U21(tt())] = [4] >= [4] = [tt()] [U31(tt(), V2)] = [2] V2 + [8] > [2] V2 + [3] = [U32(isNat(activate(V2)))] [U32(tt())] = [5] > [4] = [tt()] [U41(tt(), N)] = [7] N + [15] > [1] N + [1] = [activate(N)] [U51(tt(), M, N)] = [7] N + [7] M + [15] > [5] N + [2] M + [11] = [U52(isNat(activate(N)), activate(M), activate(N))] [U52(tt(), M, N)] = [1] N + [2] M + [12] > [1] N + [1] M + [10] = [s(plus(activate(N), activate(M)))] [s(X)] = [1] X + [4] >= [1] X + [4] = [n__s(X)] [plus(X1, X2)] = [1] X1 + [1] X2 + [4] >= [1] X1 + [1] X2 + [4] = [n__plus(X1, X2)] [U61(tt())] = [15] > [5] = [0()] [0()] = [5] > [4] = [n__0()] [U71(tt(), M, N)] = [7] N + [7] M + [15] > [6] N + [1] M + [12] = [U72(isNat(activate(N)), activate(M), activate(N))] [U72(tt(), M, N)] = [2] N + [1] M + [13] > [2] N + [1] M + [12] = [plus(x(activate(N), activate(M)), activate(N))] [x(X1, X2)] = [1] X1 + [1] X2 + [5] > [1] X1 + [1] X2 + [4] = [n__x(X1, X2)] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { U11(tt(), V2) -> U12(isNat(activate(V2))) , U12(tt()) -> tt() , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(isNat(activate(V1)), activate(V2)) , isNat(n__s(V1)) -> U21(isNat(activate(V1))) , isNat(n__x(V1, V2)) -> U31(isNat(activate(V1)), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt()) -> tt() , U31(tt(), V2) -> U32(isNat(activate(V2))) , U32(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> U52(isNat(activate(N)), activate(M), activate(N)) , U52(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> U72(isNat(activate(N)), activate(M), activate(N)) , U72(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) , x(X1, X2) -> n__x(X1, X2) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))